Tutorial Exercise Find the derivative of the function. HINT [See Example 1(b).] h(x) In[(-3x + 7)(x + 7)] Step 1 d !_[In(u)] = ²/₁ dx For the function h(x) = In[(-3x + 7)(x + 7)], note that h is a natural logarithmic function of the expression (-3x + 7)(x + 7). Hence, the generalized rule. find the derivative of h. du Note that in this case, the derivative of the expression, requires using the product rule unless the expression is expanded. To avoid expansion or using the product rule, recall the dx' logarithmic property In(a b) In(a) + In(b). If this property is applied to h, we have the following result. h(x) In[(-3x + 7)(x+7)] du will be used to dx In(-3x + 7) + In
Tutorial Exercise Find the derivative of the function. HINT [See Example 1(b).] h(x) In[(-3x + 7)(x + 7)] Step 1 d !_[In(u)] = ²/₁ dx For the function h(x) = In[(-3x + 7)(x + 7)], note that h is a natural logarithmic function of the expression (-3x + 7)(x + 7). Hence, the generalized rule. find the derivative of h. du Note that in this case, the derivative of the expression, requires using the product rule unless the expression is expanded. To avoid expansion or using the product rule, recall the dx' logarithmic property In(a b) In(a) + In(b). If this property is applied to h, we have the following result. h(x) In[(-3x + 7)(x+7)] du will be used to dx In(-3x + 7) + In
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Tutorial Exercise**
**Find the derivative of the function. HINT [See Example 1(b).]**
\[ h(x) = \ln[(-3x + 7)(x + 7)] \]
---
**Step 1**
For the function \( h(x) = \ln[(-3x + 7)(x + 7)] \), note that \( h \) is a natural logarithmic function of the expression \((-3x + 7)(x + 7)\). Hence, the generalized rule \(\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}\) will be used to find the derivative of \( h \).
Note that in this case, \(\frac{du}{dx}\), the derivative of the expression, requires using the product rule unless the expression is expanded. To avoid expansion or using the product rule, recall the logarithmic property \(\ln(a \cdot b) = \ln(a) + \ln(b)\).
If this property is applied to \( h \), we have the following result.
\( h(x) = \ln[(-3x + 7)(x + 7)] \)
\[= \ln(-3x + 7) + \ln(x + 7) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd48171ba-d1cc-47b1-bf9c-8041143384f0%2F7ef61d8b-0410-4c6f-a282-6ca5d530894d%2F4297c4f_processed.png&w=3840&q=75)
Transcribed Image Text:**Tutorial Exercise**
**Find the derivative of the function. HINT [See Example 1(b).]**
\[ h(x) = \ln[(-3x + 7)(x + 7)] \]
---
**Step 1**
For the function \( h(x) = \ln[(-3x + 7)(x + 7)] \), note that \( h \) is a natural logarithmic function of the expression \((-3x + 7)(x + 7)\). Hence, the generalized rule \(\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}\) will be used to find the derivative of \( h \).
Note that in this case, \(\frac{du}{dx}\), the derivative of the expression, requires using the product rule unless the expression is expanded. To avoid expansion or using the product rule, recall the logarithmic property \(\ln(a \cdot b) = \ln(a) + \ln(b)\).
If this property is applied to \( h \), we have the following result.
\( h(x) = \ln[(-3x + 7)(x + 7)] \)
\[= \ln(-3x + 7) + \ln(x + 7) \]
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